Comments on Prove $e^x \ge x+1 \\\; \forall x \in \mathbb{R}$ using induction
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Prove $e^x \ge x+1 \\\; \forall x \in \mathbb{R}$ using induction
(How) can we prove $e^x \ge x+1 \; \forall x \in \mathbb{R}$ using induction (without using the derivative of $e^x$ at any stage)? Comments on my attempt are appreciated.
I stumbled across a very nice proof of $\frac{\mathrm{d}}{\mathrm{d}x}e^x = e^x$ that uses the identity $e^x \ge x+1$. Briefly,
\begin{align} e^x \ge x+1 & \implies e^{-x} \ge 1-x \\ & \implies e^x \le \frac{1}{1-x} \\ & \implies x+1 \le e^x \le \frac{1}{1-x} \\ & \implies x \le e^x - 1 \le \frac{1}{1-x} - 1 \\ & \implies x \le e^x - 1 \le \frac{x}{1-x} \\ & \implies 1 \le \frac{e^x - 1}{x} \le \frac{1}{1-x} \\ & \implies \left(\lim_{x \to 0} 1 = 1\right) \le \lim_{x \to 0} \frac{e^x-1}{x} \le \left(\lim_{x \to 0} \frac{1}{1-x} = 1\right) \\ & \implies \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \\ & \implies \lim_{h \to 0} e^x \cdot \frac{e^h - 1}{h} = e^x \\ & \implies \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \\ & \implies \frac{\mathrm{d}}{\mathrm{d}x}e^x = e^x \end{align}
I now need to prove $e^x \ge x+1$ while ensuring that my argument is not circular. A lot of the proofs I came across use the derivative of $e^x$ which is not ideal. There were also references to Bernoulli's inequality which has some satisfactory proofs. Nonetheless, my first idea was induction, so I wonder if this is possible over the reals. I outline my attempt below, which I'm not very certain of.
For the base case $x = 0$,$$e^0 \ge 0 + 1 \implies \lim_{x\to0}e^x \ge \lim_{x\to 0} x + 1$$ Consider $\epsilon \ge 0 \implies e^{\epsilon} \ge 1$. We now induct as follows:
$\underline{x \in \mathbb{R}_{\ge 0}}$$$e^{\epsilon} \cdot e^x \ge e^{\epsilon}(x+1) \implies \lim_{\epsilon \to 0} e^{x + \epsilon} \ge \lim_{\epsilon \to 0} e^{\epsilon}x + \lim_{\epsilon \to 0} e^{\epsilon} \ge x + \lim_{\epsilon\to 0} \epsilon + 1 \tag{1}$$ $\underline{x \in \mathbb{R}_{< 0}}$ $$\text{Replacing} \; \epsilon \; \text{with} \; -\epsilon, \; \text{we follow the same steps as in} \; (1)$$
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| Carefree Explorer | (no comment) | Aug 21, 2022 at 08:51 |
You can't do induction on the real numbers because the real numbers aren't inductively defined. That said, it's not clear what you intend "induction on reals" to mean.
Before continuing, I'd like to make an aside on proof writing. First, as a matter of presentation, it's useful to make it clear what theorem you are attempting to prove and nice to actually conclude with deducing the stated theorem. Hence the traditional "Theorem: ... Proof: ... Q.E.D." structure. In your question, while it's clear that you ultimately want to prove $e^x \ge x+1$, it's not clear that this is what the boxed reasoning at the bottom is supposed conclude, nor does that reasoning actually conclude with that statement. This leads to the next, more fundamental point.
Proofs are not a series of statements but a series of steps of reasoning. In your first chain of implications, to be a proof, each of those implications should be explicitly justified by some reasoning. In practice, filling in simple steps of reasoning is often left to the reader, so your presentation of that chain of implications is fine enough (though there are definitely subtleties that your presentation glosses over, e.g. when we take the reciprocal of both sides, the resulting inequality is no longer true for all $x$). Nevertheless, you should generally be biased toward being more explicit about the steps of reasoning, especially when you are uncertain and/or those steps are more complicated or subtle. This helps you catch mistakes, helps others catch mistakes, and, ultimately, is just what a proof is.
For induction over the naturals, the relevant proof rule is $$\dfrac{P(0) \qquad \forall n.P(n) \implies P(n+1)}{\forall n.P(n)}$$ If you aren't familiar with this notation for our purposes here it can be read as "if we've proven the statements on the top, then this rule proves the statement on the bottom". Alternatively, we can present induction as an axiom schema: $(P(0) \land (\forall n.P(n) \implies P(n+1))) \implies \forall n. P(n)$.
To use induction, we need to state which formula we're substituting for $P$ at which point we need to prove the two statements $P(0)$ and $\forall n.P(n) \implies P(n+1)$. Having done that, we may conclude, via induction, that $\forall n. P(n)$ holds.
For your "real number induction", you'd need to state what the rule is, prove it itself or give reference to a proof, and then show how you are instantiating the rule and the resolve any proof obligations the rule entails.
For actually solving the problem, i.e. proving $e^x \ge x + 1$, it's useful to have a definition for $e^x$. For example, one definition would be that it is the unique function satisfying $Df = f$ and $f(0)=1$, but that's not too useful for your purposes. Another definition would be $e^x \equiv \sum_{n=0}^\infty x^n/n!$. This definition makes it relatively easy to see that $e^x \ge x + 1$. It's immediately obvious for $x \ge 0$. You can prove $e^{x+y}=e^x e^y$ directly from the power series, from which you can prove $e^{-x} = 1/e^x$ from which $e^x > 0$ follows for all $x\in \mathbb R$. That establishes $e^x \ge x + 1$ for $x \le -1$. The only remaining case is when $x\in(-1,0)$. For this range of $x$, we can simply directly show that $\sum_{n=2}^\infty x^n/n!$ (note the index starting point) is bounded below by $0$, e.g. with an induction over the partial sums.
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